Question

3x'2 + 5 / 2x'2 - 4 find derivatives of w.r.t x​

Answer 1

Given data:

$\quad y=3x^{2}+\dfrac{5}{2x^{2}}-4$

To find:

$\quad \dfrac{dy}{dx}$

Step-by-step explanation:

Given, $y=3x^{2}+\dfrac{5}{2x^{2}}-4$

$\Rightarrow y=3x^{2}+\dfrac{5}{2}x^{-2}-4$

Now differentiating both sides with respect to $x$, we get

$\quad \dfrac{d}{dx}(y)=\dfrac{d}{dx}(3x^{2}+\dfrac{5}{2}x^{-2}-4)$

$\Rightarrow \dfrac{dy}{dx}=\frac{d}{dx}(3x^{2})+\dfrac{d}{dx}(\dfrac{5}{2}x^{-2})-\dfrac{d}{dx}(0)$

$\Rightarrow \dfrac{dy}{dx}=3\dfrac{d}{dx}(x^{2})+\dfrac{5}{2}\dfrac{d}{dx}(x^{-2})-0$

• since $\dfrac{d}{dx}(constant)=0$

$\Rightarrow \dfrac{dy}{dx}=3\times 2\times x^{2-1}+\dfrac{5}{2}\times (-2)\times x^{-2-1}$

$\Rightarrow \dfrac{dy}{dx}=6x-5x^{-3}$

$\Rightarrow \dfrac{dy}{dx}=6x-\dfrac{5}{x^{3}}$

Answer:

The required derivative is $6x-\dfrac{5}{x^{3}}$.

Answer 2

$\textbf{Given:}$

$\mathsf{\dfrac{3x^2+5}{2x^2-4}}$

$\textbf{To find:}$

$\textsf{Derivaive of the given function}$

$\textbf{Solution:}$

$\textsf{Let}\;\;\mathsf{y=\dfrac{3x^2+5}{2x^2-4}}$

$\textsf{We apply Quotient rule of differentiation}$

$\boxed{\mathsf{\dfrac{d(\frac{u}{v})}{dx}=\dfrac{v\,\frac{du}{dx}-u\,\frac{dv}{dx}}{v^2}}}$

$\textsf{Differentiate y with respect to 'x'}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{(2x^2-4)\dfrac{d(3x^2+5)}{dx}-(3x^2+5)\dfrac{d(2x^2-4)}{dx}}{(2x^2-4)^2}}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{(2x^2-4)(6x)-(3x^2+5)(4x)}{(2x^2-4)^2}}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{12x^3-24x-12x^3-20x}{(2x^2-4)^2}}$

$\implies\boxed{\mathsf{\dfrac{dy}{dx}=\dfrac{-44x}{(2x^2-4)^2}}}$

$\textbf{Find more:}$

If y=e^2x (ax+b), then prove that d^2 y/dx^2 -4 dy/dx+4y=0

https://brainly.in/question/19674230